Question: The equation of an ellipse $E$ is $\dfrac {(x-2)^{2}}{49}+\dfrac{y^2}{36} = 1$. What are its center $(h, k)$ and its major and minor radius?
Answer: The equation of an ellipse with center $(h, k)$ is $ \dfrac{(x - h)^2}{a^2} + \dfrac{(y - k)^2}{b^2} = 1$ We can rewrite the given equation as $\dfrac{(x - 2)^2}{49} + \dfrac{(y - 0)^2}{36} = 1 $ Thus, the center $(h, k) = (2, 0)$ $49$ is bigger than $36$ so the major radius is $\sqrt{49} = 7$ and the minor radius is $\sqrt{36} = 6$.